If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, then the ratio of perimeter of rectangle and that of parallelogram is k, then k is :

Options

A. `k gt1`
B. `klt1`
C. `k=1`
D. can't be determined

Answer

To solve the problem, we need to analyze the given conditions about the rectangle and the parallelogram.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have a rectangle and a parallelogram that have the same base and height.
- Let the base of both shapes be \( A \) and the height of the rectangle be \( B \) and the height of the parallelogram be \( B' \).

2. **Calculate the Area**:
- The area of the rectangle \( A_R \) is given by:
\[ A_R = \text{Base} \times \text{Height} = A \times B \]
- The area of the parallelogram \( A_P \) is given by:
\[ A_P = \text{Base} \times \text{Height} = A \times B' \]
- Since the areas are equal, we can set them equal to each other:
\[ A \times B = A \times B' \]
- This implies:
\[ B = B' \]

3. **Perimeter Calculation**:
- The perimeter of the rectangle \( P_R \) is:
\[ P_R = 2 \times \text{Length} + 2 \times \text{Width} = 2A + 2B \]
- The perimeter of the parallelogram \( P_P \) is:
\[ P_P = 2 \times \text{Base} + 2 \times \text{Side} = 2A + 2B' \]

4. **Ratio of Perimeters**:
- Now, we need to find the ratio of the perimeter of the rectangle to the perimeter of the parallelogram:
\[ \frac{P_R}{P_P} = \frac{2A + 2B}{2A + 2B'} \]
- Simplifying this gives:
\[ \frac{P_R}{P_P} = \frac{2(A + B)}{2(A + B')} = \frac{A + B}{A + B'} \]

5. **Analyzing the Heights**:
- Since \( B' > B \) (the height of the parallelogram is greater than that of the rectangle), it follows that:
\[ A + B' > A + B \]
- Therefore, the ratio \( \frac{A + B}{A + B'} < 1 \).

6. **Conclusion**:
- Thus, the ratio \( k \) of the perimeter of the rectangle to the perimeter of the parallelogram is less than 1.

### Final Answer:
The value of \( k \) is less than 1.

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